A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.280 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.00 kg plate of food and a 0.210 kg cup of coffee. Obtain the force T exerted by the thumb and the force F exerted by the four fingers. Both forces act perpendicular to the tray, which is being held parallel to the ground.
T = N (downward)
F = N (upward)
W1= weight of the tray 0.280 kg
W2= weight of the plate 1.0 kg
W3= weight of the cup 0.210 kg
X1= center of gravity of tray is at geographical center 0.200m
X2= center of gravity of plate is at geographical center with respect to its position on tray 0.240m
X3= center of gravity of cup is at geographical center with respect to its position on tray 0.380m
Xcg= (W1X1)+(W2X2)+(W3X3)/ W1+W2+W3
Xcg= [(0.280kg)(0.200m)]+[(1.0kg)(0.240m)]+[(0.210kg)(0.380m)]/(0.280+1.0+0.210)
Xcg= (0.056+0.240+0.0798)/1.49
Xcg= 0.3758/1.49
Xcg= 0.2522m
I have found the center of gravity for the whole system which is 0.2522m. How to I proceed to answer the question?
For Further Reading
Physics please check - bobpursley, Tuesday, April 24, 2007 at 5:47pm
I don't know why you did that.
The solution is to sum the moments about the edge (or any point) and set to zero. From the description, I am not certain if the thumb or the fingers are on the edge.
You have that equation, and one other: the sum of the vertical forces is zero.
That will be enough to solve.
Physics please check - Mary, Tuesday, April 24, 2007 at 7:09pm
The thumb is on top of the tray extending 0.0600m from the edge of the tray inward and exerting T downward.
The other four finger and beneath the tray entending 0.100m from the edge of the tray inward and exeting a force F upwards.
I still don't understand how to tackle this question. I thought you had to fing the center of gravity for the tray, plate and cup then using that find T and F. Please help!!!!
No, you dont. Sum all the torques (force x distance) about the edge of the tray. and set to zero. That gives you one equation with two unknowns (fthumb, ffinger)
Then, sum the vertical forces, that gives you another equation with the same unknowns, and set to zero.
Solve for the two unknowns.
Why did the clown bring a ruler to lunch? Because he wanted to measure up to the task of balancing the lunch tray!
To tackle the problem, let's first sum all the torques about the edge of the tray and set it to zero. This gives us one equation with two unknowns, T (the force exerted by the thumb) and F (the force exerted by the four fingers).
Next, let's sum the vertical forces and set it to zero. This gives us another equation with the same unknowns, T and F.
Now, we have two equations with two unknowns. Time for some algebraic fun! Solve for T and F using the equations, and voila, you'll find the values for the forces exerted by the thumb and fingers.
Remember, don't be a clown and forget to include the units in your answer!
To answer the question, you need to find the forces T and F exerted by the thumb and fingers, respectively.
Step 1: Setting up the equations
Since the tray is being held parallel to the ground, the sum of the vertical forces is zero. This can be written as:
T + F - (W1 + W2 + W3) = 0 (Equation 1)
where W1, W2, and W3 are the weights of the tray, plate, and cup, respectively.
Step 2: Torques
Now, let's sum the torques about the edge of the tray. The torque is calculated as the force multiplied by the distance from the point of rotation (edge of the tray). For the torque to be in equilibrium, the sum must be zero.
For the thumb:
T × 0.0600m = (W1 + W2 + W3) × (0.200m + 0.240m + 0.380m - 0.0600m)
For the fingers:
F × 0.100m = (W1 + W2 + W3) × (0.200m + 0.240m - 0.100m)
Step 3: Solving the equations
Now, you have two equations: the sum of torques and the sum of vertical forces. Solve these equations simultaneously to find the unknowns T and F.
T × 0.0600m = (0.280kg + 1.0kg + 0.210kg) × (0.760m)
F × 0.100m = (0.280kg + 1.0kg + 0.210kg) × (0.340m)
Solve the above equations simultaneously to find the values of T and F.
To solve the problem, you need to sum the torques (force x distance) about the edge of the tray and set it to zero. This will give you one equation with two unknowns: T (force exerted by the thumb) and F (force exerted by the four fingers).
The torque equation is given by:
T x 0.0600m - F x 0.100m = 0
Next, you need to sum the vertical forces and set it to zero. This will give you another equation with the same unknowns:
T - F + W1 + W2 + W3 = 0
Since W1, W2, and W3 can be calculated as the weights of the tray, plate, and cup respectively, you can substitute these values into the equation:
T - F + 0.280kg + 1.00kg + 0.210kg = 0
Now you have a system of two equations with two unknowns:
T x 0.0600m - F x 0.100m = 0
T - F + 1.49kg = 0
You can solve these equations simultaneously to find the values of T and F.